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Trigo Prob to prove

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    if: [tex] (1+\sqrt{1+a})\tan{\alpha} = (1+\sqrt{1-a}) [/tex]

    Prove that: [tex] \sin{4 \alpha} = a [/tex]

    2. Relevant equations

    [tex] \cos{2\alpha} = 1-2\sin^2{\alpha} [/tex]

    [tex] \tan{\alpha} = \sqrt{\frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}}[/tex]

    3. The attempt at a solution

    We probably have to simplify to express [tex] \tan{\alpha} [/tex] in terms of [tex]\sin{4\alpha}[/tex] and hence need to use the equations given above. Trying componendo and dividendo did no good as it got too complicated.

    [tex]\sin{\alpha}+\sin{\alpha}\sqrt{1+a} = \cos{\alpha}+\cos{\alpha}\sqrt{1-a}[/tex]

    [tex](\sin{\alpha}-\cos{\alpha})^2 = (\cos{\alpha}\sqrt{1-a}-\sin{\alpha}\sqrt{1+a})^2 [/tex]


    Though I could simplify after this, I cannot understand how to get [tex]\sin{4\alpha} [/tex] out of it. What is looking even more difficult is how to remove the [tex] \sqrt{1-a^2} [/tex] term.

    I would be grateful if somebody could guide me in solving this.
  2. jcsd
  3. Nov 19, 2007 #2
    Wait, is the [tex] \sqrt{1-a^2} [/tex] inside of your last sine term or is it a scalar multiplying your last sine term?
  4. Nov 20, 2007 #3
    No, it is a scalar multiplying the last sine term.
    Last edited: Nov 20, 2007
  5. Nov 20, 2007 #4
    Looks like I got it; beginning from where I left,

    [tex]1-\sin{2\alpha} = \cos^2{\alpha-a\cos^2{\alpha}+\sin^2{\alpha}+a\sin^2{\alpha}-(\sin{2\alpha})(\sqrt{1-a^2})[/tex]

    -After taking 'a' common...

    [tex]Since, \cos^2{\alpha}+\sin^2{\alpha}=1[/tex]




    the term [tex]sin^2{2\alpha}[/tex] gets cancelled on both sides and taking 'a' common we get>>

    [tex]\Rightarrow \sin{4\alpha}=a[/tex]

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